Answer
$$\frac{1}{4}{e^{{t^4}}} + C $$
Work Step by Step
$$\eqalign{
& \int {{t^3}{e^{\left( {{t^4}} \right)}}} dt \cr
& {\text{set }}u = {t^4}{\text{ then }}\frac{{du}}{{dt}} = 4{t^3} \to \frac{{du}}{{4{t^3}}} = dt \cr
& {\text{write the integrand in terms of }}u \cr
& \int {{t^3}{e^{\left( {{t^4}} \right)}}} dt = \int {{t^3}{e^u}} \left( {\frac{{du}}{{4{t^3}}}} \right) \cr
& {\text{cancel common terms}} \cr
& = \int {{e^u}} \left( {\frac{{du}}{4}} \right) = \frac{1}{4}\int {{e^u}} du \cr
& {\text{integrating}} \cr
& = \frac{1}{4}{e^u} + C \cr
& {\text{replace }} - {t^2}{\text{ for }}u \cr
& = \frac{1}{4}{e^{{t^4}}} + C \cr} $$
